If #sin(theta) = 3/5# and #theta# is in Q1, then what is #sin(90^@+theta)# ?

1 Answer
George C.
Apr 9, 2017

#4/5#

Explanation:

Note that:

#cos^2 theta + sin^2 theta = 1#

So:

#cos theta = +-sqrt(1-sin^2 theta)#

#color(white)(cos theta) = +-sqrt(1-(3/5)^2)#

#color(white)(cos theta) = +-sqrt((25-9)/25)#

#color(white)(cos theta) = +-sqrt(16/25)#

#color(white)(cos theta) = +-4/5#

We can identify the correct sign as #+# since #theta# is in Q1.

So:

#cos theta = 4/5#

The sum of angles formula for #sin# tells us:

#sin(alpha+beta) = sin(alpha)cos(beta)+sin(beta)cos(alpha)#

Putting #alpha = 90^@# and #beta = theta# we find:

#sin(90^@+theta) = sin(90^@)cos(theta)+sin(theta)cos(90^@)#

#sin(90^@+theta) = 1*cos(theta)+sin(theta)*0#

#sin(90^@+theta) = cos(theta)#

#sin(90^@+theta) = 4/5#

Note that in passing we have shown:

#sin(90^@+theta) = cos(theta)#

for any #theta#.

That is, #cos(theta)# is the same as #sin(theta)# shifted by #90^@#.

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